Question

Determine the number of grams of Na2SO4 needed to prepare 350.0 mL of a solution which is 0.500M Na2SO4. ___________g Na2SO4 (b) A titration was used to determine the amount of Cl-1 ion in a water supply, using a reaction with a standardized solution containing Ag+1 as follows: Ag+1 (aq) + Cl-1(aq) → AgCl (s) A 10.0-g water sample required 20.2 mL of 0.100 M Ag+1 for complete reaction. Calculate the grams of Cl-1 and the mass percent chloride ion present. _______ g Cl-1 _______ % Cl-1

Answer 1

Answer:

a) 24.85 grams of sodium sulfate is needed.

b) Mass of 0.00202 moles of chloride ions:

   Mass percentage of chloride ion present in the sample is 0.7171%.

Explanation:

$M=\frac{n}{V(L)}$

m = Molarity of the solution

n = moles of compound

V = volume of the solution in L.

a) Moles of sodium sulfate = n

Molarity of the solution , M= 0.500 M

Volume of the solution = V = 350.0 mL = 0.3500 L

$n=M\times V=0.500 M\times 0.3500 L=0.175 mol$

Mass of 0.175 moles of sodium sulfate = 0.175 mol × 142 g/mol = 24.85 g

24.85 grams of sodium sulfate is needed.

b) Moles of silver ion = n

Molarity of the silver ions = M = 0.100 M

Volume of the solution = V = 20.2 mL = 0.0202 L

$n=M\times V= 0.100 M\times 0.0202 L=0.00202 mol$

$Ag^++Cl^-\rightarrow AgCl$

According to reaction , 1 mole of silver ions reacts with 1 mole of chloride ions.

Then 0.00202 moles of silver ions will react with :

$\frac{1}{1}\times 0.00202 mol=0.00202 mol$ of chloride ions.

Mass of 0.00202 moles of chloride ions:

0.00202 mol × 35.5 g/mol = 0.07171 g

Mass percentage of chloride ions in 10.0 grams of water:

$=\frac{0.07171 g}{10.0 g}\times 100=0.7171\%$

Mass percentage of chloride ion present in the sample is 0.7171%.